The radii of internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm, respectively.

Solution:

We have,

the internal base radius of spherical shell, $r_{1}=3 \mathrm{~cm}$,

the external base radius of spherical shell, $r_{2}=5 \mathrm{~cm}$ and

the base radius of solid cylinder, $r=\frac{14}{2}=7 \mathrm{~cm}$

Let the height of the cylinder be $h$.

As,

Volume of solid cylinder = Volume of spherical shell

$\Rightarrow \pi r^{2} h=\frac{4}{3} \pi r_{2}^{3}-\frac{4}{3} \pi r_{1}^{3}$

$\Rightarrow \pi r^{2} h=\frac{4}{3} \pi\left(r_{2}^{3}-r_{1}^{3}\right)$

$\Rightarrow r^{2} h=\frac{4}{3}\left(r_{2}^{3}-r_{1}^{3}\right)$

$\Rightarrow 7 \times 7 \times h=\frac{4}{3}\left(5^{3}-3^{3}\right)$

$\Rightarrow 49 \times h=\frac{4}{3}(125-27)$

$\Rightarrow h=\frac{4}{3} \times \frac{98}{49}$

$\therefore h=\frac{8}{3} \mathrm{~cm}$

So, the height of the cylinder is $\frac{8}{3} \mathrm{~cm}$.